Answer to Question #107271 in Differential Equations for Nitin Chandra gahtori

Question

d²y/dx²-2tanx=0, dy/DX+5y=e^xsecx

Answer 1

$1)\frac{d^2y}{dx^2}-2\tan x=0\\ y''-2\tan x=0\\ y''=2\tan x\\ \frac{dy'}{dx}=2\frac{\sin x}{\cos x}\\ y'=2\int\frac{\sin x}{\cos x}dx\\ y'=-2ln|\cos x|+c_1\\ \frac{dy}{dx}=-2ln|\cos x|+c_1\\ y=\int(-2ln|\cos x|+c_1)dx\\ y=-2xln|\cos x|-2\int x\tan x dx+xc_1+c_2$

$2) \frac{dy}{dx}+5y=\frac{e^x}{\cos x}\\ y'+5y=\frac{e^x}{\cos x}\\ i) y'+5y=0\\ \frac{dy}{dx}=-5y\\ \frac{dy}{y}=-5dx\\ ln|y|=-5x+ln|c|\\ y_0=c\cdot e^{-5x}\\ ii) y=c(x)\cdot e^{-5x}\\ c'e^{-5x}-5ce^{-5x}+5ce^{-5x}=\frac{e^x}{\cos x}\\ c'=\frac{e^{6x}}{\cos x}\\ c=\int \frac{e^{6x}}{\cos x}dx+c_1\\ y=(\int \frac{e^{6x}}{\cos x}dx+c_1)\cdot e^{-5x}$